Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
mt19937 rd(time(0));
#define int long long
const int M=1004535809;
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
const int N=250005;
pii f1[N],f2[N];
char s[N];
int n;
map<pair<int,int>,int> Map;
int ksm(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=x*x%M)
		if (y&1)(ans*=x)%=M;
	return ans;
}
pii fuhe(pii x,pii y){
	return mp((x.fi*y.fi)%M,(x.fi*y.se+x.se)%M);
}
signed main(){
	scanf("%lld",&n);
	scanf("%s",s+1);
	int x1=rd()%(M-1)+1,x2=rd()%(M-1)+1;
	int Inv1=ksm(x1,M-2),Inv2=ksm(x2,M-2);
	f1[0]=mp(1,0);f2[0]=mp(1,0);
	for (int i=1;i<=n;i++){
		if (s[i]=='+')f1[i]=fuhe(f1[i-1],mp(1,1)),f2[i]=fuhe(f2[i-1],mp(1,1));
		if (s[i]=='-')f1[i]=fuhe(f1[i-1],mp(1,M-1)),f2[i]=fuhe(f2[i-1],mp(1,M-1));
		if (s[i]=='<')f1[i]=fuhe(f1[i-1],mp(x1,0)),f2[i]=fuhe(f2[i-1],mp(x2,0));
		if (s[i]=='>')f1[i]=fuhe(f1[i-1],mp(Inv1,0)),f2[i]=fuhe(f2[i-1],mp(Inv2,0));
	}
	int ans=0;
	for (int i=1;i<=n;i++){
		Map[mp((fuhe(f1[i-1],f1[n])).se,(fuhe(f2[i-1],f2[n])).se)]++;
		ans+=Map[mp(f1[i].se,f2[i].se)];
	}
	printf("%lld\n",ans);
}